Mister Exam
Differential equation x^3y'''-3x^2y''+6xy'-6y=3+ln(x^3)
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Solve with variation of parameters
Solve x^3 (d^3 y(x))/(dx^3) - 3 x^2 (d^2 y(x))/(dx^2) + 6 x (dy(x))/(dx) - 6 y(x) = log(x^3) + 3:
The general solution will be the sum of the complementary solution and particular solution.
Find the complementary solution by solving x^3 (d^3 y(x))/(dx^3) - 3 x^2 (d^2 y(x))/(dx^2) + 6 x (dy(x))/(dx) - 6 y(x) = 0:
Assume a solution to this Euler-Cauchy equation will be proportional to x^λ for some constant λ.
Substitute y(x) = x^λ into the differential equation:
x^3 (d^3 )/(dx^3) (x^λ) - 3 x^2 (d^2 )/(dx^2) (x^λ) + 6 x d/(dx) (x^λ) - 6 x^λ = 0
Substitute (d^3 )/(dx^3) (x^λ) = λ (λ - 2) (λ - 1) x^(λ - 3), (d^2 )/(dx^2) (x^λ) = λ (λ - 1) x^(λ - 2), and d/(dx) (x^λ) = λ x^(λ - 1):
λ^3 x^λ - 6 λ^2 x^λ + 11 λ x^λ - 6 x^λ = 0
Factor out x^λ:
(λ^3 - 6 λ^2 + 11 λ - 6) x^λ = 0
Assuming x!=0, the zeros must come from the polynomial:
λ^3 - 6 λ^2 + 11 λ - 6 = 0
Factor:
(λ - 3) (λ - 2) (λ - 1) = 0
Solve for λ:
λ = 1 or λ = 2 or λ = 3
The root λ = 1 gives y_1(x) = c_1 x as a solution, where c_1 is an arbitrary constant.
The root λ = 2 gives y_2(x) = c_2 x^2 as a solution, where c_2 is an arbitrary constant.
The root λ = 3 gives y_3(x) = c_3 x^3 as a solution, where c_3 is an arbitrary constant.
The general solution is the sum of the above solutions:
y(x) = y_1(x) + y_2(x) + y_3(x) = c_1 x + c_2 x^2 + c_3 x^3
Determine the particular solution to x^3 (d^3 y(x))/(dx^3) - 3 x^2 (d^2 y(x))/(dx^2) + 6 x (dy(x))/(dx) - 6 y(x) = 3 log(x) + 3 by variation of parameters:
List the basis solutions in y_c(x):
y_(b_1)(x) = x, y_(b_2)(x) = x^2, and y_(b_3)(x) = x^3
Compute the Wronskian matrix of y_(b_1)(x), y_(b_2)(x), and y_(b_3)(x):
W_M(x) = (
x | x^2 | x^3
d/(dx) (x) | d/(dx) (x^2) | d/(dx) (x^3)
(d^2 )/(dx^2) (x) | (d^2 )/(dx^2) (x^2) | (d^2 )/(dx^2) (x^3)) = (
x | x^2 | x^3
1 | 2 x | 3 x^2
0 | 2 | 6 x)
Compute the Wronskian determinant:
W(x) = left bracketing bar W_M(x) right bracketing bar = left bracketing bar
x | x^2 | x^3
1 | 2 x | 3 x^2
0 | 2 | 6 x right bracketing bar = 2 x^3
Compute the following determinants:
W_1(x) | = left bracketing bar
0 | x^2 | x^3
0 | 2 x | 3 x^2
1 | 2 | 6 x right bracketing bar = x^4
| where W_1(x) is the determinant of W_M(x) with the first column replaced by (0 | 0 | 1)^T
W_2(x) | = left bracketing bar
x | 0 | x^3
1 | 0 | 3 x^2
0 | 1 | 6 x right bracketing bar = -2 x^3
| where W_2(x) is the determinant of W_M(x) with the second column replaced by (0 | 0 | 1)^T
W_3(x) | = left bracketing bar
x | x^2 | 0
1 | 2 x | 0
0 | 2 | 1 right bracketing bar = x^2
| where W_3(x) is the determinant of W_M(x) with the third column replaced by (0 | 0 | 1)^T
Divide the differential equation by the leading term's coefficient x^3:
(d^3 y(x))/(dx^3) - (3 (d^2 y(x))/(dx^2))/x + (6 (dy(x))/(dx))/x^2 - (6 y(x))/x^3 = (3 log(x) + 3)/x^3
Let f(x) = (3 log(x) + 3)/x^3:
Let v_1(x) = integral(f(x) W_1(x))/W(x) dx, v_2(x) = integral(f(x) W_2(x))/W(x) dx, and v_3(x) = integral(f(x) W_3(x))/W(x) dx:
The particular solution will be given by:
y_p(x) = v_1(x) y_(b_1)(x) + v_2(x) y_(b_2)(x) + v_3(x) y_(b_3)(x)
Compute v_1(x):
v_1(x) = integral(3 log(x) + 3)/(2 x^2) dx = -3/x - (3 log(x))/(2 x)
Compute v_2(x):
v_2(x) = integral-(3 log(x) + 3)/x^3 dx = 9/(4 x^2) + (3 log(x))/(2 x^2)
Compute v_3(x):
v_3(x) = integral(3 log(x) + 3)/(2 x^4) dx = -2/(3 x^3) - log(x)/(2 x^3)
The particular solution is thus:
y_p(x) = v_1(x) y_(b_1)(x) + v_2(x) y_(b_2)(x) + v_3(x) y_(b_3)(x) = x (-3/x - (3 log(x))/(2 x)) + x^2 (9/(4 x^2) + (3 log(x))/(2 x^2)) + x^3 (-2/(3 x^3) - log(x)/(2 x^3))
Simplify:
y_p(x) = 1/12 (-6 log(x) - 17)
The general solution is given by:
Answer: |
| y(x) = y_c(x) + y_p(x) = c_1 x + c_2 x^2 + c_3 x^3 + 1/12 (-6 log(x) - 17)
Solve with a substitution
Solve x^3 (d^3 y(x))/(dx^3) - 3 x^2 (d^2 y(x))/(dx^2) + 6 x (dy(x))/(dx) - 6 y(x) = log(x^3) + 3:
Let y(x) = x v(x), which gives (dy(x))/(dx) = x (dv(x))/(dx) + v(x), (d^2 y(x))/(dx^2) = x (d^2 v(x))/(dx^2) + 2 (dv(x))/(dx), and (d^3 y(x))/(dx^3) = x (d^3 v(x))/(dx^3) + 3 (d^2 v(x))/(dx^2):
6 x (x (dv(x))/(dx) + v(x)) - 3 x^2 (x (d^2 v(x))/(dx^2) + 2 (dv(x))/(dx)) + x^3 (x (d^3 v(x))/(dx^3) + 3 (d^2 v(x))/(dx^2)) - 6 x v(x) = log(x^3) + 3
Simplify:
x^4 (d^3 v(x))/(dx^3) = log(x^3) + 3
Divide both sides by x^4:
(d^3 v(x))/(dx^3) = 3/x^4 + (3 log(x))/x^4
Integrate both sides with respect to x:
(d^2 v(x))/(dx^2) = integral(3/x^4 + (3 log(x))/x^4) dx = -4/(3 x^3) - log(x)/x^3 + c_1, where c_1 is an arbitrary constant.
Integrate both sides with respect to x:
(dv(x))/(dx) = integral(-4/(3 x^3) - log(x)/x^3 + c_1) dx = 11/(12 x^2) + log(x)/(2 x^2) + x c_1 + c_2, where c_2 is an arbitrary constant.
Integrate both sides with respect to x:
v(x) = integral(11/(12 x^2) + log(x)/(2 x^2) + c_1 x + c_2) dx = -17/(12 x) - log(x)/(2 x) + (x^2 c_1)/2 + x c_2 + c_3, where c_3 is an arbitrary constant.
Simplify the arbitrary constants:
v(x) = -17/(12 x) - log(x)/(2 x) + c_1 x^2 + c_2 x + c_3
Substitute back for y(x) = x v(x), which gives v(x) = y(x)/x:
y(x)/x = -17/(12 x) - log(x)/(2 x) + c_1 x^2 + c_2 x + c_3
Solve for y(x):
y(x) = 1/12 (-6 log(x) + 12 c_1 x^3 + 12 c_2 x^2 + 12 c_3 x - 17)
Simplify the arbitrary constants:
Answer: |
| y(x) = -log(x)/2 + c_1 x^3 + c_2 x^2 + c_3 x - 17/12
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